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Unformatted text preview: Boltzmanno number of 1particle state >>number of particles How many 1particle states? Remember the sphere used to explain degeneracy? number of 1particle states with an energy lower than = number of lattice points enclosed by the sphere in the positive 3 octant: ( ) 1 4 8 1 m = R3 = 8 3 6 h 2 3 a for thermal particles, 3 = = kT per particle 2
For the Bolztman condition to hold,
3 2 ( ) 1 m 3kT 8 = V 6 h 2 N 1 m 3kT V 8 N ? 1 6 h 2 m
3 2 3 2 ( ) ? 6 h which will occur for T =1 mkT 12 an d to express the energy of N particles,
1 ln Q ( N,V ,T ) Q ( N ,V , T ) E =  =  Q ( N ,V , T ) V V
N  k 1 e N ! k E = 1 q ( N,V ,T ) N! N N  k 1 1 N  k E = e k (  k ) e N N q ( N,V ,T ) ! k N!  k E = ke q ( N,V ,T ) k N E =N The average energy of N particles is N times the average energy of one particle e  k in terms of the probability , = k k where k = ,  e j k # of available states ? N o we have to count more carefully we have to look at the particle's wavefunction Fermions: 2 particles cannot occupy the same quantum stateooccupation i1 i2 i3
k number 1,2 can only be 0 or 1 n P = antisymmetric Bosons: 2 particles can occupy the same quantum stateooccupation number nk can have any value P1,2=+ symmetric i2 i2 i3 i4 i5 i6 There are many ways of counting distributions, 1 fluctuations of physical observables o X for large systems, fluctuations are negligible o all ensembles are equivalent choose GrandCanonical Ensemble ( V ,T , ) = e N j  ENJ e  N ENj = energy states available for a system of N particles nk ( E j ) = molecules in k quantum state, in the E j state k = molecular quantum state { nk } defines the system at E j E j = k k and N = k n n e j E NJ = e  i i i n k k a sum over states becomes the sum over each distribution { }
ni ( V ,T , ) = N N e j  ENJ where we used = e

i i i  if we consider = e j E NJ = e
{ n} n a sum over states becomes a sume over each distribution N =0 e n
N  n i i i where
n m m { } n k k =N = e N =0  n i i i { n} (  ( n11 +n2 2 +...+nk k +...) ( n1 +n2 +...+nk +...) ) = e N =0 { n} = N =0 { n} k ( e  k nk ) = n1=nmax n1=nmax
1 n L n =0
2 =0 k ( e  k nk ) where we use the fact that N can have any value (0 ) ( V,T, ) = nk =nmax
k nk =0 ( e  k nk ) starting with ( V,T, ) = nk =nmax
k nk =0 ( e  k nk ) we look at Fermions, w here nk =0,1 ( V, T , ) = =
k nk =1
k nk =0 ( e  k nk )  k e ( e ) +( ) 0  k 1 ( V ,T , ) = k ( 1+ e  k ) FermiDirac again, we start with ( V,T, ) = we now look at Bosons , where ( V, T , ) =
nk =
k nk =nmax
k nk =0 ( e  k nk ) nk = 0,1,...N nk =0 ( e  k nk ) from our outstanding math background ... we recognize the series 1 x = 1  x m =0
o m ( V ,T , ) = 1
k 1 e  k BoseEinstein Combining BoseEinstein with the FermiDirac we obtain a gral. Eq. ( V ,T , ) =
+ + e  k )  1 ( 1 k for a Grandcanonical ensemble, N = kt and using = e kT ln ( V ,T , ) 1 kT d = e d kT ln ( V ,T , ) 1 ln ( V ,T , ) N = kt = kT + ln ( 1+ e  k ) e  k k N = = and N = + e  k ) k (1 nk
k for each quantum state, the average number of particles nk = ( 1+ e  k )  e  k e  k which is equivalent to j = e  k k we can now calculate all other thermo properties: k e  k E = N k = nk k =  k + ) k k ( 1 e and PV = kT ln ( V ,T , ) = kT
k ln ( 1+ e  k )  Keep in mind that even though we started with noninteracting particles, there wavefunctions are symmetrized othey are NOT INDEPENDENT to see if these eq. hold for Bolztmann statistics conditions Boltzmanno number of 1particle state >>number of particles o nk 0, which will occur when 0 for 0 0 nk e  k = 1 e  k nk = e  k 1+ e  k ) ( nk N
k e  k = k (Boltzmannclassical limit) e  k and for the energy, E = lim E = k e  k 0 k k e  k k +  ( 1 e k )
k E N = e = e k k
k k k e = e  k k
k k =
finally,
k k k (Boltzmannclassical limit)
+ + e  k )  1 =+ kT ( 1 pV = kt ln ( V ,T , ) = kT ln for pV = kT k 0 we can expand ln ( 1 x ) x + e  k = kt q(N,V ,T ) pV ln ( V ,T , ) = kt k k ln ( 1+ e  k )  ( V ,T , ) = e q ( N,V ,T ) Examples: Photon gas, an electromagnetic field in thermal equilibrium with its container QM we know that H =
i hi In=thisncase hi =harmonic oscillator i h To describe the state of the field, we need to know how many n are in each oscillator Photons are bosons o n=0,1,2,3..... Q ( N,V,T ) =  E = e n1 ,n2 ,...,n j e  ( n11 + n22 +...+ n j j ) first we want to show that Q ( N,V,T ) =
n1 ,n2 ,...,n j e  ( n11 ) e  ( n22 ) L e  ( n j j ) =
j n j =0 e ( n j j ) Q ( N,V,T ) = ( e  ( 01 ) + e  ( 11 ) +  ( 21 ) ) e ) ( e  ( 0 +
3 = e  ( 01 + 02 ) + e  ( 01 +12 ) + e  ( 01 + 22 ) + e  ( 11 + 02 ) + e  ( 11 +12 ) + e  ( 11 + 22 ) + e  ( 21 + 02 ) + e  ( 21 +12 ) + e  ( 11 + 22 ) ) ( e  ( 0 +
3 ) e  ( 13 + e  ( 23 ) ) ( e  ( 0 )
2 + e  ( 12 ) + e  ( 22 ) ) ) e  ( 13 + e  ( 23 ) ) = e  ( 01 + 02 + 0 3 ) + e  ( 01 + 02 +13 ) + e  ( 01 + 02 + 23 ) + e  ( 11 + 02 )
1 2 3 1 2 +e  ( 01 +12 + 03 ) + e  ( 01 +12 +13 ) + e  ( 0 +1 + 2 ) + e  ( 0 + 2 + 0 3 ) 1 2 3 1 2 3 1 2 +e  ( 0 + 2 +1 ) + e  ( 0 + 2 + 2 ) + e  ( 1 + 0 + 03 ) + ... e  ( n11 + n22 + n33 )
n1 ,n2 ,...,n j 1 2 3 = e  ( 01 + 02 + 0 3 ) + e  ( 11 + 0 2 + 03 ) + e  ( 0 +1 + 0 ) 2 3 1 2 3 1 2 + e  ( 0 1 + 0 +1 ) + e  ( 2 + 0 + 0 ) + e  ( 0 + 2 + 0 3 ) + ... Q ( N,V,T ) = j e  ( nj j ) n j =0 =  ( nj j ) j 1 e 1 where we used
xj =
j 1 1 x The average number of photons in a state j...  E n j e  ( n11 + n22 +...) nje nj = = e  E { n} Q ln Q ( N,V ,T )
nj 1 nj = Q { n} n j e  ( n11 + n22 +...)  j ) n (
j =
nj  ( j ) n ln Q ( N,V ,T ) = (  1) ln ( 1  e  j ) ln Q ( N,V ,T )  j ) ( 1 e + j e  j = 1 e  j  average occupation number is n j = = Planck  1 distribution ...
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This note was uploaded on 05/29/2011 for the course CHM 6461 taught by Professor Bowers during the Spring '08 term at University of Florida.
 Spring '08
 Bowers
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